Given Q = -6.0 nC, a = 30 cm, and b = 40cm, find: If Q = -6.0 nC, a = 30cm, and

Question

Given Q = -6.0 nC, a = 30 cm, and b = 40cm, find:

If Q = -6.0 nC, a = 30cm, and b = 40 cm in the figure, Write an expression for r in terms of a and b pointing from the charge Q on top to point P? Draw r on the diagram. Write an expression for r in terms of a and b pointing from the charge Q on bottom to point P? Draw r on the diagram. What is the electric field at point P in terms of a and b, and the unit vectors in terms of a and b? Draw the electric field on the diagram. What is the electric potential at point P? Assume V(lemniscate) = 0. How much work is needed to move a charge + 2Q point P to point 0? Governing Equations: integral E vector d A vector = Q enc/epsilon_0, phi_E = integral E vector d A vector

Explanation / Answer

a) r= sqrt(a^2+b^2)

b) r= sqrt(a^2+b^2)

c)

r= sqrt(a^2+b^2) = sqrt(30^2+40^2)= 50

= tan^-1(30/50) = 30.96

E= kQ/r^2 = (9*10^9*6.0*10^-9)/(0.50^2) = 216 N/C

From the above figure you will find y components of both E cancels each other and Enet is sum of two x components.

Thus Enet = 2Ex = 2*(216)*cos30.96 = 370.45 N/C

d) Vnet(P) = V1 + V2 = kQ/r^2 + kQ/r^2 = 2kQ/r^2 = (2*9*10^9*-6.0*10^-9)/0.5 = -216 V

e) Vnet(Q) = V1 + V2 = kQ/r^2 + kQ/r^2 = 2kQ/r^2 = (2*9*10^9*-6.0*10^-9)/0.3 = -360 V

W= q*V = (2Q)*[ Vnet(P) - Vnet(Q)] = (2*-6.0*10^-9)*(-360-(-216)) = 1.73*10^-6 J

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